Answer to Question #246527 in Statistics and Probability for lucy

Question

The monthly saving of 500 students is normally distributed with mean RM200 and standard deviation RM150.

a) find the number of students has saving between RM180 and RM281 a month.

b) determine the value of X given that 85% of the students saving more than RMx a month.

Answer 1

Given :

Sample size

n = 500

Mean

μ = 200

Standard deviation

σ = 0

a)

P(180 < x < 281)

Using z-score formula ,

$z =\frac{x-μ}{σ}\\ =P(\frac{180 – 200}{150}< \frac{x-μ}{σ}< \frac{281 – 200}{150})\\ = P(-0.13 < z < 0.54)\\ = P (< 0.54) - P(Z <-0.13)$

Using Excel function, =NORMSDIST (z)

=NORMSDIST (0.54), = 0.7054

=NORMSDIST (-0.13), = 0.4483

=0.7054 - 0.4483

=0.2571

P(180 < x < 281) = 0.2571

500*0.2571 = 128.55

Rounding to nearest whole number would be 129

Therefore the number of students has saving between $ 180 and $ 281 a month are 129

That is 129 students has saving between $ 180 and $ 281 a month.

b)

P(X>x)=0.85

P(X<x) =1-0.85

P(X<x)=0.15

Using Excel function, =NORMSINV (area)

=NORMSINV (0.15)

=-1.04 (rounded to two decimals)

z = -1.04

Using the z-score formula ,

$z=\frac{x-μ}{σ}$

x= μ+zσ

x= 200+ (-1.04 * 150)

x= 200+ (-156)

x= 44